A374015 Residue modulo 5 of n! divided by the highest power of 10 which divides n!.

1, 1, 2, 1, 4, 2, 2, 4, 2, 3, 3, 3, 1, 3, 2, 3, 3, 1, 3, 2, 4, 4, 3, 4, 1, 4, 4, 3, 4, 1, 3, 3, 1, 3, 2, 2, 2, 4, 2, 3, 2, 2, 4, 2, 3, 1, 1, 2, 1, 4, 2, 2, 4, 2, 3, 4, 4, 3, 4, 1, 1, 1, 2, 1, 4, 1, 1, 2, 1, 4, 3, 3, 1, 3, 2, 4, 4, 3, 4, 1, 3, 3, 1, 3, 2, 2, 2, 4, 2, 3, 2, 2, 4, 2, 3, 1, 1, 2, 1, 4, 4, 4, 3, 4, 1, 3

Offset: 0

Author

Cezary Glowacz, Jun 25 2024

Keywords

• nonn
• easy

Comments

The sequence is not eventually periodic. This because by induction on k the eventual period must be a multiple of 5^k for every k.

a(5^k) = 2^k mod 5.

From Cezary Glowacz, Feb 07 2025: (Start)

The proportions p(m,s) of counts of pairs of consecutive terms s among a(1),...,a(m) converge to equidistribution (and as an immediate consequence, so do proportions of individual terms).

This can be seen, for example, by stating p(5^(4(n+1)+1)-1,s) as affine functions of p(5^(4n+1)-1,t) and examining the convergence of p(5^(4n+1)-1,u) to the equidistribution. Then, p(m,s) converges to the equidistribution because the maximum over s of the absolute values of deviations from 1/16 of p(m,s) for m>k*5^(4n+1)-1 is less than the corresponding maximum over t for p(5^(4n+1)-1,t) plus 2/(5^(4n+1)) + 1/k.

Consecutive terms 1,2,3 do not occur, so that triples do not have a similar equidistribution.

(End)

If n > 0 is not divisible by 5, a(n) == n * a(n-1) (mod 5). - Robert Israel, Jul 05 2024

Examples

			a(5) = 1*2*3*4*5/10 mod 5 = 2.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..20000

Crossrefs

Cf. A004154, A010874, A145679.

Cf. A045549, A045547, A045550, A045548.

Programs

Maple

a:= n-> (f-> irem(f/10^padic[ordp](f, 10), 5))(n!):
seq(a(n), n=0..105);  # Alois P. Heinz, Jun 25 2024

Mathematica

a[n_]:=Mod[n!/10^IntegerExponent[n!, 10],5]; Array[a,106,0] (* Stefano Spezia, Jun 25 2024 *)

PARI

a(n)=if(n>4, my(k=n\5); return(lift((n%5)!*a(k)*Mod(2,5)^k))); n!%5 \\ Charles R Greathouse IV, Jan 24 2025

Python

v=[[((1,1,2,1,4)[j]*2**(i*j))%5 for j in range(5)] for i in range(4)]
def a(n):
    c,p=0,1
    while n: c,n,p=(c+1)%4,n//5,(v[c][n%5]*p)%5
    return(p) # Cezary Glowacz, Feb 05 2025

Formula

a(n) = A010874(A004154(n)).